
#1
May1106, 10:46 PM

P: 77

I have to calculate this without using the calculator.
cos (arctan 5/12) So far i draw a triangle and i have the opposite side to be 5, adjacent to be 12, and hypotenuse to be 13. Please suggest me some hint, thanks. 



#2
May1106, 10:52 PM

Mentor
P: 7,292

Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan [itex] \frac 5 {12} [/itex]
then use the definition of the cos of that angle to get your answer. 



#3
May1106, 11:23 PM

P: 77





#4
May1106, 11:30 PM

HW Helper
P: 2,875

Trig problem 



#5
May1206, 08:36 AM

P: 77

Thanks, should the unit be the length of the sides, since 12/13 is not an angle. 



#6
May1206, 08:37 AM

HW Helper
P: 2,875





#7
May1306, 09:16 AM

P: 602

Just for another take on this problem (I find the construction of a rt triangle a bit cumbersome) we could use basic trig. results.
For eg, if you have to do something like cos(arctan(x)) we can proceed by taking arctan(x)=y so x=tan(y) [tex]\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+tan^2y}} = cos(y) or y=arccos(\frac{1}{\sqrt{1+x^2}})[/tex] So cos(arctan(x)) = cos(y) = [itex]\frac{1}{\sqrt{1+x^2}})[/itex] which gives the answer.This approach works for all such problems. No messy triangles. Arun 



#8
May1306, 10:04 PM

P: 77

Thanks everyone for helping me out.



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